With absolute values, a sketch usually helps . . .
We know the graph of . . . a V with vertex at the origin.
The graph of . . . "climbs steeper".
The graph of . . . inverts the previous graph.
The graph of . . . moved up one unit.
-2 /::|::\ 2
- - + - o - + - o - + - -
|::/ | \::|
|:/ | \:|
|/ | \|
* | *
We have an inverted V, vertex at (0,1), x-intercepts at
They ask for the "area" from to
. . and some of it is "negative area".
We don't need Calculus to find the areas of triangles.
[My answer is: -8]